1. **Stating the problem:** A smooth sphere weighing $10\sqrt{3}$ grams rests against a smooth vertical wall. It is suspended by a string attached at a point on its surface. We need to analyze the forces acting on the sphere and find the tension in the string or other related quantities. 2. **Understanding the setup:** Since the sphere is smooth and rests against a smooth vertical wall, the wall exerts a normal reaction force horizontally. The string exerts a tension force at the point of suspension on the sphere's surface. 3. **Forces involved:** - Weight $W = 10\sqrt{3}$ grams (acting vertically downward). - Tension $T$ in the string (direction depends on the string's orientation). - Normal reaction $N$ from the wall (horizontal, perpendicular to the wall). 4. **Equilibrium conditions:** Since the sphere is at rest, the net force in both horizontal and vertical directions must be zero. 5. **Using geometry:** Let the radius of the sphere be $r$. The string is attached at the surface, so the point of suspension is at a distance $r$ from the center. 6. **Resolving forces:** - Vertically: $T \sin \theta = W$ - Horizontally: $N = T \cos \theta$ where $\theta$ is the angle the string makes with the vertical. 7. **Finding tension:** From vertical equilibrium, $$T = \frac{W}{\sin \theta}$$ 8. **Additional relations:** The geometry of the sphere and string attachment point can give $\theta$ if more data is provided. Since the problem statement is incomplete, this is the general approach to solve it. **Final answer:** The tension in the string is given by $$T = \frac{10\sqrt{3}}{\sin \theta}$$ where $\theta$ is the angle between the string and the vertical.